Problem: As you very well know, this year's funkiest numbers are so called triangular numbers (that is, integers that are representable as <image>, where k is some positive integer), and the coolest numbers are those that are representable as a sum of two triangular numbers.

A well-known hipster Andrew adores everything funky and cool but unfortunately, he isn't good at maths. Given number n, help him define whether this number can be represented by a sum of two triangular numbers (not necessarily different)!

Input

The first input line contains an integer n (1 ≤ n ≤ 109).

Output

Print "YES" (without the quotes), if n can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes).

Examples

Input

256


Output

YES


Input

512


Output

NO

Note

In the first sample number <image>.

In the second sample number 512 can not be represented as a sum of two triangular numbers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long nx = 2e5 + 5;
const long long inf = 1e18 + 7;
const long long mod = 1e9 + 7;
vector<long long> adj[nx];
vector<long long> a;
bool bs(long long val, long long s, long long e) {
  if (e >= s) {
    long long mid = (s + e) / 2;
    if (a[mid] == val) {
      return true;
    }
    if (val > a[mid]) {
      s = mid + 1;
      return bs(val, s, e);
    }
    if (val < a[mid]) {
      e = mid - 1;
      return bs(val, s, e);
    }
  }
  return false;
}
signed main() {
  long long n;
  cin >> n;
  for (long long i = 1; i <= nx; i++) {
    a.push_back(i * (i + 1) / 2);
    if (a[i - 1] >= 1e9) break;
  }
  long long l = a.size();
  bool k = false;
  for (long long i = 0; i < l; i++) {
    long long need = n - a[i];
    if (need > 0) {
      k = bs(need, 0, l - 1);
      if (k == true) break;
    }
  }
  if (k == true) cout << "YES" << endl;
  if (k == false) cout << "NO" << endl;
  return 0;
}